Amity University Noida B.Tech Admissions 2025
ApplyAmong Top 30 National Universities for Engineering (NIRF 2024)
19 Questions around this concept.
The solution for of the equation is
We have already learned to find Indefinite Integration by using the substitution method. But in the case of definite integration, we also need to change the limits of integration 'a' and 'b'. If we substitute x = g(t), then g(t) must be continuous in the interval [a, b].
Let's look at some examples of how such questions are solved.
Example 1
Compute the integral $\int_0^{\pi / 2} \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}$
Let $\quad I=\int_{x=0}^{x=\pi / 2} \frac{d x}{a^2 \cos ^2 x+b^2 \sin ^2 x}$
Divide numerator and denominator by $\cos ^2 x$
$
=\int_{x=0}^{x=\pi / 2} \frac{\sec ^2 x d x}{a^2+b^2 \tan ^2 x}
$
Put $\quad \tan x=t \Rightarrow \sec ^2 x d x=d t$
$
\therefore \quad I=\int_{t=0}^{t=\infty} \frac{d t}{a^2+b^2 t^2}
$
We find the new limits of integration $t=\tan x \Rightarrow t=0$ when $x=0$ and $t=\infty$ when $x=\pi / 2$
$
\begin{aligned}
\Rightarrow \quad I & =\frac{1}{b^2} \int_0^{\infty} \frac{d t}{\left(\frac{a}{b}\right)^2+t^2}=\frac{1}{b^2} \cdot \frac{1}{a / b}\left[\tan ^{-1} \frac{b t}{a}\right]_0^{\infty} \\
& =\frac{1}{a b}\left[\frac{\pi}{2}-0\right]=\frac{\pi}{2 a b}
\end{aligned}
$
"Stay in the loop. Receive exam news, study resources, and expert advice!"